RD Sharma solution class 7 chapter 2 Fractions Exercise 2.2

Exercise 2.2

Page-2.19

Question 1:

Multiply:
(i) $\frac{7}{11} \mathrm{by} \frac{3}{5}$
(ii) $\frac{3}{5} \mathrm{by} 25$
(iii) $3\frac{4}{15} \mathrm{by} 24$
(iv) $3\frac{1}{8} \mathrm{by} \times 4\frac{10}{14}$

Answer 1:

(i)
$$\begin{gathered} \frac{7}{11}\times \frac{3}{5}=\frac{7\times 3}{11\times 5} \\ \Rightarrow \frac{21}{55} \end{gathered}$$

(ii)
$$\begin{gathered} \frac{3}{5}\times 25=\frac{3\times 25}{5\times 1} \\ \Rightarrow \frac{75}{5}\iff 15 \end{gathered}$$

(iii)
$$\begin{gathered} \left[3\frac{4}{15}\right]\times 24 \\ \left[\frac{(3\times 15)+4}{15}\right]\times 24=\frac{49\times 24}{15\times 1} \\ \Rightarrow \frac{392}{5}= 78\frac{2}{5} \end{gathered}$$

(iv)
$$\begin{gathered} \left[3\frac{1}{8}\right]\times \left[4\frac{10}{11}\right] \\ \left[\frac{(3\times 8)+1}{8}\right]\times \left[\frac{(4\times 11)+10}{11}\right]=\frac{25\times 54}{8\times 11} \\ \Rightarrow \frac{1350}{88}\iff \frac{675}{44} = 15\frac{15}{44} \end{gathered}$$

Question 2:

Find the product:
(i) $\frac{4}{7}\times \frac{14}{25}$
(ii) $7\frac{1}{2}\times 2\frac{4}{15}$
(iii) $3\frac{6}{7}\times 4\frac{2}{3}$
(iv) $6\frac{11}{14}\times 3\frac{1}{2}$

Answer 2:

(i)
$$\begin{gathered} \frac{4}{7}\times \frac{14}{25}=\frac{4\times 14}{7\times 25} \\ \Rightarrow \frac{8}{25} \end{gathered}$$

(ii)
$$\begin{gathered} 7\frac{1}{2}\times 2\frac{4}{15}\iff \frac{(7\times 2)+1}{2}\times \frac{(2\times 15)+4}{15} \\ \Rightarrow \frac{15}{2}\times \frac{34}{15}\iff \frac{17}{1} \end{gathered}$$
(iii)
$$\begin{gathered} 3\frac{6}{7}\times 4\frac{2}{3}\iff \frac{(3\times 7)+6}{7}\times \frac{(4\times 3)+2}{3} \\ \Rightarrow \frac{{\cancel{27}}^9}{\cancel{7}}\times \frac{{\cancel{14}}^2}{\cancel{3}}\iff \frac{18}{1} \end{gathered}$$
(iv)
$$\begin{gathered} 6\frac{11}{14}\times 3\frac{1}{2}\iff \frac{(14\times 6)+11}{14}\times \frac{(3\times 2)+1}{2} \\ \Rightarrow \frac{95}{\cancel{{14}^2}}\times \frac{\cancel{7}}{2}\iff \frac{95}{4} \end{gathered}$$

Page-2.20

Question 3:

Simplify:
(i) $\frac{12}{25}\times \frac{15}{28}\times \frac{35}{36}$
(ii) $\frac{10}{27}\times \frac{39}{56}\times \frac{28}{65}$
(iii) $2\frac{2}{17}\times 7\frac{2}{9}\times 1\frac{33}{52}$

Answer 3:

(i)
$$\begin{gathered} \frac{12}{25}\times \frac{15}{28}\times \frac{35}{36}\iff \frac{\cancel{12}\times {\cancel{15}}^3\times {\cancel{35}}^5}{{\cancel{25}}^5\times {\cancel{28}}^4\times {\cancel{36}}^3} \\ \Rightarrow \frac{\cancel{3}\times \cancel{5}}{\cancel{5}\times 4\times \cancel{3}}\iff \frac{1}{4} \end{gathered}$$

(ii)
$$\begin{gathered} \frac{10}{27}\times \frac{39}{56}\times \frac{28}{65}\iff \frac{\cancel{10}\times {\cancel{39}}^{{\cancel{3}}^1}\times \cancel{28}}{{\cancel{27}}^9\times {\cancel{56}}^{\cancel{2}}\times {\cancel{65}}^{\cancel{5}}} \\ \Rightarrow \frac{1}{9} \end{gathered}$$
(iii)
$$\begin{gathered} 2\frac{2}{17}\times 7\frac{2}{9}\times 1\frac{33}{52}\iff \frac{(2\times 17)+2}{17}\times \frac{(7\times 9)+2}{9}\times \frac{(52\times 1)+33}{52} \\ \iff \frac{\cancel{{36}^{\cancel{4}}}\times {\cancel{65}}^5\times {\cancel{85}}^5}{\cancel{17}\times \cancel{9}\times {\cancel{52}}^{\cancel{4}}} \\ \Rightarrow \frac{25}{1} \end{gathered}$$

Question 4:

Find:
(i) $\frac{1}{2} \mathrm{of} 4\frac{2}{9}$
(ii) $\frac{5}{8} \mathrm{of} 9\frac{2}{3}$
(iii) $\frac{2}{3} \mathrm{of} \frac{9}{16}$

Answer 4:

$$\begin{gathered} \left(\mathrm{i}\right) \frac{1}{2} \mathrm{of} 4\frac{2}{9} \\ =\frac{1}{2}\times \left(4\frac{2}{9}\right) \\ =\frac{1}{2}\times \frac{(4\times 9)+2}{9}=\frac{1\times 38}{2\times 9} \\ =\frac{19}{9}=2\frac{1}{9} \\ \left(\mathrm{ii}\right) \frac{5}{8} \mathrm{of} 9\frac{2}{3} \\ =\frac{5}{8}\times \left(9\frac{2}{3}\right) \\ =\frac{5}{8}\times \frac{(9\times 3)+2}{3}=\frac{5\times 29}{8\times 3} \\ =\frac{145}{24}=6\frac{1}{24} \\ \left(\mathrm{iii}\right) \frac{2}{3} \mathrm{of} \frac{9}{16}=\frac{2\times 9}{3\times 16} \\ =\frac{3}{8} \end{gathered}$$

Question 5:

Which is greater? $\frac{1}{2} \mathrm{of} \frac{6}{7} \mathrm{or} \frac{2}{3} \mathrm{of} \frac{3}{7}$

Answer 5:

$$\begin{gathered} \frac{1}{2}\mathrm{of} \frac{6}{7}=\frac{1}{2}\times \frac{6}{7}=\frac{6}{14} \\ \frac{2}{3}\mathrm{of} \frac{3}{7}=\frac{2}{3}\times \frac{3}{7}=\frac{2}{7} \end{gathered}$$
$\mathrm{c}\mathrm{onvert} \frac{2}{7}\mathrm{to} \mathrm{its} \mathrm{equivalent} \mathrm{fraction} \mathrm{with} \mathrm{denominator} \mathrm{as} 14$
$\frac{2}{7}=\frac{2}{7}\times \frac{2}{2}=\frac{4}{14}$
$we \mathrm{know}6>4$
$\Rightarrow \frac{6}{14}>\frac{2}{7}$
$\Rightarrow \frac{1}{2}\mathrm{of} \frac{6}{7}>\frac{2}{3}\mathrm{of} \frac{3}{7}$

Question 6:

Find:
(i) $\frac{7}{11}$ of Rs 330
(ii) $\frac{5}{9}$ of 108 metres
(iii) $\frac{3}{7}$ of 42 litres
(iv) $\frac{1}{12}$ of an hour
(v) $\frac{5}{6}$ of an year
(vi) $\frac{3}{20}$ of a kg
(vii) $\frac{7}{20}$ of a litre
(viii) $\frac{5}{6}$ of a day
(ix) $\frac{2}{7}$ of a week

Answer 6:

$$\begin{gathered} \left(i\right) \frac{7}{11}of Rs 330=\frac{7\times 330}{11} \\ \Rightarrow Rs 210 \\ \left(ii\right) \frac{5}{9}of 108 metres=\frac{5}{9}\times 108 \\ \Rightarrow 60 metres \\ \left(iii\right) \frac{3}{7}of 42 litres=\frac{3}{7}\times 42 \\ \Rightarrow 18 litres \\ \left(iv\right) \frac{1}{12}of 1 hour=\frac{1}{12}\times 1 \\ \Rightarrow \frac{1}{12}hour \\ 1 hour =60 minutes \\ \Rightarrow \frac{1}{12}hour=\frac{1}{12}\times 60=5 minutes \\ \left(v\right) \frac{5}{6}of 1 year=\frac{5}{6}\times 1 \\ \Rightarrow \frac{5}{6}year \\ 1 year =12 months \\ \Rightarrow \frac{5}{6}year=\frac{5}{6}\times 12=10 months \\ \left(vi\right) \frac{3}{20}of 1 kg=\frac{3}{20}\times 1 \\ \Rightarrow \frac{3}{20}kg \\ 1 kg =1000 g \\ \Rightarrow \frac{3}{20}kg=\frac{3}{20}\times 1000=150 g \\ \left(vii\right) \frac{7}{20}of 1 litre=\frac{7}{20}\times 1 \\ \Rightarrow \frac{7}{20}litre \\ 1 l =1000 ml \\ \Rightarrow \frac{7}{20}litre=\frac{7}{20}\times 1000=350 ml \\ \left(viii\right) \frac{5}{6}of 1 day=\frac{5}{6}\times 1 \\ \Rightarrow \frac{5}{6}day \\ 1 day =24 hours \\ \Rightarrow \frac{5}{6}day=\frac{5}{6}\times 24=20 hours \\ \left(ix\right) \frac{2}{7}of 1 week=\frac{2}{7}\times 1 \\ \Rightarrow \frac{2}{7}week \\ 1 week =7 days \\ \Rightarrow \frac{2}{7}week=\frac{2}{7}\times 7=2 days \end{gathered}$$

Question 7:

Shikha plants 5 saplings in a row in her garden. The distance between two adjacent saplings is $\frac{3}{4}\mathrm{m}$. Find the distance between the first and the last sapling.

Answer 7:

Distance between the first and second saplings = $\frac{3}{4}\mathrm{m}$
Distance between the first and third saplings = $2\times \frac{3}{4}\mathrm{m}=\frac{3}{2}\mathrm{m}$
Distance between the first and fourth saplings = $3\times \frac{3}{4}\mathrm{m}=\frac{9}{4}\mathrm{m}$

Distance between the first and fifth saplings  = 4 × $\frac{3}{4}\mathrm{m}=3 \mathrm{m}$

Question 8:

Ravish reads $\frac{1}{3}$ part of a book in 1 hour. How much part of the book will he read in $2\frac{1}{5}$ houurs?

Answer 8:

$$\begin{gathered} 2\frac{1}{5}\mathrm{hours}=\frac{(2\times 5)+1}{5}=\frac{11}{5}\mathrm{hours} \\ \mathrm{In} 1 \mathrm{hour} \mathrm{Ravish} \mathrm{reads} \frac{1}{3} \mathrm{of} \mathrm{the} \mathrm{book} \\ \mathrm{Part} \mathrm{of} \mathrm{book} \mathrm{Ravish} \mathrm{will} \mathrm{read} \mathrm{in} \frac{11}{5}\mathrm{hours} = \text{P}\mathrm{art} \mathrm{read} \mathrm{in} 1 \mathrm{hour} \times \frac{11}{5} \\ \mathrm{Part} \mathrm{of} \mathrm{book} \mathrm{Ravish} \mathrm{will} \mathrm{read} \mathrm{in} \frac{11}{5}\mathrm{hours} =\frac{1}{3}\times \frac{11}{5}=\frac{11}{15} \end{gathered}$$

Question 9:

Lipika reads a book for $1\frac{3}{4}$ hours everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book?

Answer 9:

In one day, Lipika reads for $1\frac{3}{4}\mathrm{hours}$.
$1\frac{3}{4}\mathrm{hours}=\frac{(1\times 4)+3}{4}=\frac{7}{4}\mathrm{hours}$
Total hours required = $\frac{7}{4}\times 6=\frac{42}{4}\mathrm{hours} = \frac{21}{2}\mathrm{hours} = 10\frac{1}{2}\mathrm{hours}$

Question 10:

Find the area of a rectangular park which is $41\frac{2}{3} \mathrm{m}$ long and $18\frac{3}{5} \mathrm{m}$ broad.

Answer 10:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{rectangular} \mathrm{park} = 41\frac{2}{3}\mathrm{m} =\frac{(41\times 3)+2}{3}\mathrm{m} \\ \Rightarrow \mathrm{Length} \mathrm{of} \mathrm{rectangular} \mathrm{park} = \frac{125}{3}\mathrm{m} \\ \mathrm{Width} \mathrm{of} \mathrm{rectangular} \mathrm{park} = 18\frac{3}{5}\mathrm{m} =\frac{(18\times 5)+3}{5}\mathrm{m} \\ \Rightarrow \mathrm{Width} \mathrm{of} \mathrm{rectangular} \mathrm{park} = \frac{93}{5}\mathrm{m} \\ \mathrm{Area} \mathrm{of} \mathrm{rectangular} \mathrm{park} = \mathrm{Length} \mathrm{of} \mathrm{rectangular} \mathrm{park} \times \mathrm{Width} \mathrm{of} \mathrm{rectangular} \mathrm{park} \\ \mathrm{Area} \mathrm{of} \mathrm{rectangular} \mathrm{park} = \frac{125}{3}\times \frac{93}{5}=775{\mathrm{m}}^2 \end{gathered}$$

Question 11:

If milk is available at Rs $17\frac{3}{4}$ per litre, find the cost of $7\frac{2}{5}$ litres of milk.

Answer 11:

Cost of 1 litre milk = $\text{Rs}. 17\frac{3}{4}$
$17\frac{3}{4}=\frac{(17\times 4)+3}{4}=\frac{71}{4}$
$\Rightarrow \mathrm{Cost} \mathrm{of} 1 \mathrm{litre} \mathrm{milk} =\text{Rs}. \frac{71}{4}$
$$\begin{gathered} 7\frac{2}{5}\mathrm{litres}=\frac{(7\times 5)+2}{5}=\frac{37}{5}\mathrm{litres} \\ \mathrm{Cost} \mathrm{of} \frac{37}{5}\mathrm{litres} \mathrm{milk} = \text{C}\mathrm{ost} \mathrm{of} 1 \mathrm{litre} \mathrm{milk} \times \frac{37}{5} \\ \mathrm{Cost} \mathrm{of} \frac{37}{5}\mathrm{litres} \mathrm{milk}=\frac{71}{4}\times \frac{37}{5}=\mathrm{Rs}. \frac{2627}{20}= \mathrm{Rs}. 131\frac{7}{20} \end{gathered}$$

Question 12:

Sharda can walk $8\frac{1}{3} \mathrm{km}$ in one hour. How much distance will she cover in $2\frac{2}{5}$ hours?

Answer 12:

$$\begin{gathered} 8\frac{1}{3}\mathrm{km}=\frac{(8\times 3)+1}{3} \\ \Rightarrow 8\frac{1}{3}\mathrm{km}=\frac{25}{3}\mathrm{km} \end{gathered}$$

Distance covered by ​Sharda in 1 hour =  $\frac{25}{3}$km
$2\frac{2}{5}\mathrm{hours} =\frac{(2\times 5)+2}{5}=\frac{12}{5}\mathrm{hours}$
Distance covered by Sharda in $\frac{12}{5}\mathrm{hours}$ = Distance covered in 1 hour $\times \frac{12}{5}$
$\Rightarrow$Distance covered by Sharda in $\frac{12}{5}\mathrm{hours}$ = $\frac{25}{3}\times \frac{12}{5}=20 \mathrm{km}$

Question 13:

A sugar bag contains 30 kg of sugar. After consuming $\frac{2}{3}$ of it, how much sugar is left in the bag?

Answer 13:

$$\begin{gathered} \mathrm{Total} \mathrm{amount} \mathrm{of} \mathrm{sugar} \mathrm{in} \mathrm{the} \mathrm{bag} = 30 \mathrm{kg} \\ \frac{2}{3} \mathrm{of} 30 \mathrm{kg} =\frac{2}{3}\times 30 = 20 \text{kg} \\ \Rightarrow \mathrm{Amount} \mathrm{of} \mathrm{sugar} \mathrm{consumed} = 20 \mathrm{kg} \\ \mathrm{Sugar} \mathrm{left} \mathrm{in} \mathrm{the} \mathrm{bag} = \mathrm{Total} \mathrm{amount} \mathrm{of} \mathrm{sugar} \mathrm{in} \mathrm{the} \mathrm{bag} - \mathrm{Amount} \mathrm{of} \mathrm{sugar} \mathrm{consumed} \\ \mathrm{Sugar} \mathrm{left} \mathrm{in} \mathrm{the} \mathrm{bag} = 30-20 =10 \mathrm{kg} \end{gathered}$$

Question 14:

Each side of a square is $6\frac{2}{3} \mathrm{m}$ long. Find its area.

Answer 14:

$$\begin{gathered} 6\frac{2}{3}\mathrm{m} = \frac{(6\times 3)+2}{3}\mathrm{m} \\ \Rightarrow 6\frac{2}{3}\mathrm{m} = \frac{20}{3}\mathrm{m} \\ \mathrm{Area} \mathrm{of} \text{s}\mathrm{quare}=\mathrm{Side} \times \mathrm{Side} \\ \mathrm{Area} \mathrm{of} \text{s}\mathrm{quare}=\frac{20}{3} \times \frac{20}{3} \\ \Rightarrow \mathrm{Area} \mathrm{of} \text{s}\mathrm{quare}=\frac{400}{9}{\mathrm{m}}^2 =44\frac{4}{9}{\text{m}}^2 \end{gathered}$$

Question 15:

There are 45 students in a class and $\frac{3}{5}$ of them are boys. How many girls are there in the class?

Answer 15:

$$\begin{gathered} \frac{3}{5} \mathrm{of} 45 =\frac{3}{5}\times 45 =27 \\ \mathrm{Therefore}, 27 \mathrm{students} \mathrm{are} \mathrm{boys}. \\ \mathrm{Number} \mathrm{of} \mathrm{girls}=\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{students} \mathrm{in} \mathrm{the} \mathrm{class} - \mathrm{Number} \mathrm{of} \mathrm{boys} \\ \mathrm{Number} \mathrm{of} \mathrm{girls}=45-27=18 \\ \mathrm{Therefore}, 18 \mathrm{students} \mathrm{are} \mathrm{girls}. \end{gathered}$$